# Hooke's law and angular freq.

misogynisticfeminist
hmm ok, i was watching the MIT opencourseware video on oscillations and there was a part where it was mentioned that,

the diff. eq. $$x''+ \frac {k}{m} x = 0$$ has solution $$x= x_0 cos (\omega t + \phi)$$ if and only if $$\omega= \sqrt{\frac {k}{m}}$$

how do i show that omega is the sqaureoot of k over m? thanks alot.

Homework Helper
Differentiate the solution twice with respect to time and substitute it back into the differential equation - then tell us what you discovered! :)

misogynisticfeminist
ohhh, i see, it can be gotten by verifying the solution. I thought we needed to do something to hooke's law, but verifying the solution works great too.

: )

edit: a tinge of doubt crosses my mind though. When we solve the original differential equation, we get the solution in terms of k, m and x and no omega. While verifying the solution works when the solution is given, and we see that $$\omega= \sqrt{\frac {k}{m}}$$. How do we know that $$\omega= \sqrt{\frac {k}{m}}$$ when we are solving it?

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Homework Helper
If you just substitute $x = x_0 \cos(\omega t - \phi)$ into the differential equation the required value of $\omega$ will jump out at you!

Staff Emeritus
edit: a tinge of doubt crosses my mind though. When we solve the original differential equation, we get the solution in terms of k, m and x and no omega. While verifying the solution works when the solution is given, and we see that $$\omega= \sqrt{\frac {k}{m}}$$. How do we know that $$\omega= \sqrt{\frac {k}{m}}$$ when we are solving it?
In this case, $$\omega$$ is referring to the angular frequency, which is the multiplicative factor in front of the independent variable in the sine or cosine function. It's just a standard substitution that they probably just didn't bother to define.